Find the number of arrangements of $k \mbox{ }1'$s, $k \mbox{ }2'$s, $\cdots, k \mbox{ }n'$s - total $kn$ cards - so that
no same numbers appear consecutively. For $k=2$ we can compute it by using the
PIE, and it is $$\frac{1}{2^n} \sum_{i=0}^n (-1)^i \binom{n}{i} (2n-i)! 2^i$$

1 Answer
1

I believe the answer is given by $$\int_0^\infty e^{-x} q_k(x)^n \, dx$$ where $q_k(x) = \sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$ for $k\geq 1$, and $q_0(x) = 1$. In general if we allow $k_i$ of the $i$th number the answer should be $$\int_0^\infty e^{-x} \prod_i q_{k_i}(x) \, dx$$

You can check that this agrees with the sequences oeis.org/A114938 and oeis.org/A193638 above. I do not (quite) have a proof of this, although I'm very close. The method is my own, and has not been published anywhere as far as I know. I'd be happy to give you more information in private, but I'm not sure I want to expose it publicly until it's proven. Please let me know if you think this is noteworthy and any potential applications. the

Edit: Following some information given to me by Byron, I found that this formula is already known and that in fact $q_n(x) = (-1)^{n}L_n^{(-1)}(x)$ where $L_n^{(\alpha)} (x) $ denotes the generalized Laguerre polynomial. See Section 6 here for a labelled version. I should have mentioned this sooner; thanks Byron!